In this article, we shall learn the methods on How To Find Series Resistance. The calculation for equivalent resistance in a series circuit is relatively straightforward and does not require complex mathematics.

**Suppose we have two resistors, R _{1} and R_{2}, as shown in image 1. We know that the current remains the same at every point of a series circuit and potential drops at each resistor.**

**Therefore, [Latex] V_{1} = iR_{1} [/Latex] and [Latex] V_{2} = iR_{2} [/Latex].**

**The total voltage in the circuit is [Latex] V= V_{1} + V_{2} = iR_{1} + iR_{2} [/Latex]. **

**If R is the equivalent resistance, then [Latex] V= iR [/Latex] **

**So, [Latex] iR= iR_{1} + iR_{2} [/Latex] and [Latex] R= R_{1} + R_{2} [/Latex].**

**How To Find Series Resistance – FAQs**

**What are the characteristics of series resistance?**

Series resistances have quite a few features in a circuit, out of which the most significant one is – the equivalent resistance is the simple addition of all the resistors joined via a series connection.

**The other series resistance characteristics are-**

**The current passing through each resistor joined in series is equal.****The voltage drop through a series resistor depends upon the value of that resistor, and it is equal to the current × resistance value.****The equivalent resistance in series is always greater than the individual resistances.**

**How does series resistance affect current?**

Unlike parallel resistances, the current remains unaffected when it passes through series resistors. It is the ratio of the source voltage and the equivalent resistance, i.e., the sum of the resistors.

**The current flow in a circuit requires some resistance to be present in the circuit. The current at each resistor joined in series is equal. As there is no branch in the case of series connection, the current does not undergo any split. Therefore, we get the same current everywhere in the circuit, which is the total current. **

Read more on….How To Calculate Voltage In A Series Circuit: Detailed Facts

**How To Find Series Resistance- FAQs**

**What is the voltage drop in a series circuit?**

From ohm’s law, we understand voltage drop in a series circuit. It is the decline of voltage when current passes through the conductor and is measured between two end points lying along the conductor.

**For any resistor sharing the series connection, the voltage drop depends linearly upon the value of the resistance. The voltage drop between the two sides of the resistor = value of the resistor × current flowing through the series circuit. The more the resistance, the more the amount of voltage drops.**

Read more on…How To Calculate Voltage Drop In A Series Circuit: Detailed Facts

**What is the significance of series resistance?**

In any circuit, resistance controls and limits the current flow. An imbalance of resistance can cause an open circuit (if the resistance is very small) or a short circuit (if the resistance is very high), leading to damage.

**Resistors in a series circuit are often called “current limiters’ as they restrict the current flow. For example, in a light emitting diode (LED), we limit the current that gets through the LED to protect it from overheating. The series resistor limits the current so that the LED can blow up without damage.**

Also read…Is Voltage Constant In Series: Complete Insights and FAQs

**How To Find Series Resistance- Numerical problems**

**A. Compute the following values for the circuit shown in image **

**1.Equivalent series resistance **

**2. The current through each resistor**

**3.The voltage drop across each resistor**

In the circuit, we can see three resistors joined in series. Therefore equivalent series resistance = R_{1}+ R_{2} + R_{3} = 2+3+5 = 10 ohm.

We know, total current i = source voltage/ equivalent resistance = 25/10= 2.5 A

**Now, the voltage drop across any resistor in series = total current in the series circuit * resistance of that resistor. **

**Therefore, the voltage drop across the 2 ohm resistor= 2.5 * 2 = 5 V**

**The voltage drop across the 3 ohm resistor= 2.5 * 3 = 7.5 V**

**The voltage drop across the 5 ohm resistor= 2.5 * 5 = 12.5 V **

**B. Image 3 depicts a circuit with four resistors where R**_{1}= 3 R_{2}, R_{2} = 2 R_{3}, and R_{3}= 5R_{4}. The supply voltage is 18 V. A current of 2 mA passes through the circuit. Find the values of the resistors and equivalent series resistance.

_{1}= 3 R

_{2}, R

_{2}= 2 R

_{3}, and R

_{3}= 5R

_{4}. The supply voltage is 18 V. A current of 2 mA passes through the circuit. Find the values of the resistors and equivalent series resistance.

Equivalent resistance for the circuit [Latex] = R_{1} + R_{2} + R_{3} + R_{4} = 3 \times 2 \times 5\: R_{4} + 2 \times 5\: R_{4} + 5\: R_{4} = 45R_{4}\; \Omega [/Latex]

By ohm’s law, V=iR or R= V/i.

So, equivalent resistance = supply voltage / current [Latex] = \frac{18} {2 \times 10^{-3}} = 9 \times 10^{3} \: \Omega [/Latex]

**Now, we can calculate the values of the resistors from the relations given. **

**Therefore, [Latex] 45 R_{4} = 9000\; or R_{4} = 200\; \Omega [/Latex]**

**[Latex] R_{3} = 5R_{4} = 5 \times 200 = 1000\; \Omega\; or\; 1\; k\Omega [/Latex]**

**[Latex] R_{2} = 2R_{3} = 2 \times 1000 = 2000\; \Omega\; or\; 2\; k\Omega [/Latex]**

**[Latex] R_{1} = 3R_{2} = 3 \times 2000 = 6000\; \Omega\; or\; 6\; k\Omega [/Latex]**

**Equivalent series resistance = R _{1} + R_{2} + R_{3} + R_{4} = 200+ 1000+ 2000+ 6000 = 9200 ohm or 9.2 kohm.**